# -*- coding:utf-8 -*-


# Given a sorted array nums, remove the duplicates in-place such that each element appear only once and return the new length.
#
# Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.
#
# Example 1:
#
#
# Given nums = [1,1,2],
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# Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively.
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# It doesn't matter what you leave beyond the returned length.
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# Example 2:
#
#
# Given nums = [0,0,1,1,1,2,2,3,3,4],
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# Your function should return length = 5, with the first five elements of nums being modified to 0, 1, 2, 3, and 4 respectively.
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# It doesn't matter what values are set beyond the returned length.
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#
# Clarification:
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# Confused why the returned value is an integer but your answer is an array?
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# Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.
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# Internally you can think of this:
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#
# // nums is passed in by reference. (i.e., without making a copy)
# int len = removeDuplicates(nums);
#
# // any modification to nums in your function would be known by the caller.
# // using the length returned by your function, it prints the first len elements.
# for (int i = 0; i < len; i++) {
#     print(nums[i]);
# }
#


class Solution(object):
    def removeDuplicates(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if not nums:
            return 0
        n = len(nums)
        count = curr = 1
        while count < n:
            if nums[curr] != nums[curr-1]:
                curr += 1
            else:
                del nums[curr]
            count += 1
        return curr
